Slips and Sums

Today's question is a pure thinking question. Its answer is very short, but it's a great question nonetheless. Read it carefully, then try to answer. Don't look at the answer just yet, it's easy when you think.

A jar is filled with a thousand paper slips which have the numbers 1 - 1000. Each time two slips are taken out of the jar, and the difference between the numbers on them is written on a new slip of paper. The difference is put back in the jar and the two original numbers are taken out. Will the number on the last slip of paper be odd or even?

Like many other questions, this one looks impossible at first. There are so many papers, and it's hard to keep track on them. Trying out all possibilities will take way too long, but luckily, using a very powerful tool, your brain, will solve this question easily.

Try looking at the sum of all numbers in the jar. At first, it's 500,500 (using the sum formula of natural numbers from 1 to n: n(n + 1)/2). Then you take out two numbers and put back their difference. However, the parity does not change.

If the sum of two numbers is odd, their difference is also odd. The same goes for even sums. When you take out two numbers and put back their difference, the sum goes down, but the parity does not change. You either take out an even sum and put back an even number, or take out an odd sum and put back an odd number.

Since the parity of the sum of the numbers in the jar is always kept even (because the initial sum is even), the last two slips will have an even sum. Because their sum is even, their difference is also even, and the number on the last slip will be even.

Hope you liked it,
Nadav

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